1. **State the problem:** Kriti has 20 coins consisting of 1-rupee, 2-rupee, and 5-rupee coins totaling 41 rupees. The number of 2-rupee coins is three times the number of 5-rupee coins. We need to find the number of 1-rupee coins. 2. **Define variables:** Let the number of 5-rupee coins be $x$. Then the number of 2-rupee coins is $3x$ (three times the 5-rupee coins). Let the number of 1-rupee coins be $y$. 3. **Write equations:** Total coins: $$x + 3x + y = 20$$ Simplify: $$4x + y = 20$$ Total value: $$5x + 2(3x) + 1 imes y = 41$$ Simplify: $$5x + 6x + y = 41$$ $$11x + y = 41$$ 4. **Solve the system:** From the first equation: $$y = 20 - 4x$$ Substitute into the second equation: $$11x + (20 - 4x) = 41$$ $$11x + 20 - 4x = 41$$ $$7x + 20 = 41$$ $$7x = 21$$ $$x = 3$$ 5. **Find $y$:** $$y = 20 - 4(3) = 20 - 12 = 8$$ 6. **Answer:** Kriti has **8** one-rupee coins.